FINAL ROUND

Let, by condition, $$n=29q_1+r_1=41q_2+r_2=59q_3+r_3=r_1+r_2+r_3,$$ $r_1<29$, $r_2<41$, $r_3<59$. From these inequalities it follows that $59q_3=r_1+r_2\le 28+40=68$, so $q_3=1$. Then $$r_1+r_2=59.(1)$$

Further, $41q_2=r_1+r_3\le 28+58=86$, so $q_2\le 2$. Consider two cases: $$1)q_2=1.\text{Then}{r}_1+r_3=41.(2)$$ From (1) and (2) it follows that $$100=59+41=r_1+r_2+r_1+r_3=n+r_1=29q_1+2r_1,$$ i.e., $29q_1+2r_1=100$. Hence $q_1$ is even and $29q_1<100$, so $q_1=2$. Then $r_1=\frac{1}{2}(100-2\cdot 29)=21$, $r_2=38$, $r_3=20$, and $n=21+38+20=79$, which satisfies the problem condition.

2) $q_2=2$. Then $$r_1+r_3=n-r_2=41q_2=82.(3)$$ From (1) and (3) it follows that $29q_1+2r_1=141$, so $q_1$ is odd and $29q_1<141$. Hence either $q_1=1$ or $q_1=3$. If $q_1=1$, then $2r_1=122$, i.e., $r_1=61$, a contradiction. If $q_1=3$, then $r_1=27$, $r_2=32$, $r_3=55$, and $n=27+32+55=114$, which satisfies the problem condition.